\(\left(x-2\right)^2=\left(2x+3\right)^2\)

\(\frac{4}{2x-3}-\frac{7}{3x-5}=6\)

\(\frac{2}{2x+1}+\frac{x}{4x^2-1}=\frac{7}{2x-1}\)

giải phương trình sau :

a/ x + 1 = 3x - 1

b/x - 4 +2x = 4 - x

c/\(\frac{x+3}{4}+1=\frac{1-x}{6}\)

\(\frac{2x+3}{4}=1+\frac{x-7}{5}\)

\(^{\left(x+1\right)^2-\left(x-1\right)^2=2x\left(x+1\right)-6}\)

\(\left(1+x\right)^3-\left(x-2\right)^2=9\left(x-1\right)^2\)